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3 years ago

Which explains the information needed to calculate speed and velocity?

Physics

1 answer:

Anestetic [448]3 years ago

3 0

Both require time, but velocity requires displacement and speed requires distance.

Explanation:

Speed and velocity are two different quantities. In fact:

- Speed is a scalar quantity, which gives a measure of how fast an object is moving, regardless of its direction. Therefore, it only has a magnitude, which is given by the ratio between distance covered and time taken:

$s=\frac{distance}{time}$

Therefore, speed does not take into account the direction of motion.

- Velocity, on the other hand, is a vector quantity, so it has magnitude and a direction.

The magnitude of the velocity is given by

$v=\frac{displacement}{time}$

where displacement is a vector connecting the initial point with the final point of motion of an object.

The direction of the velocity corresponds to the direction of the displacement.

It must be noted also that in certain situations, the average velocity is zero, while the speed is not zero: for example, for an object completing a circle in a certain time interval, its speed is not zero, while its velocity is zero (because the object returns to the starting point, so the displacement is zero.

Therefore, the correct answer is

Both require time, but velocity requires displacement and speed requires distance.

Learn more about speed and velocity:

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Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant